// Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.

// Return the intersection of these two interval lists.

// (Formally, a closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b. 
//  The intersection of two closed intervals is a set of real numbers that is either empty, 
//  or can be represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].)


// Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
// Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
// Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.


USE TWO POINTERS

//TC: O(n + m) because we just loop through length of intervals
//SC: O(1) because algorithm doesnt use any extra space and output array doesnt count
//IF WE ARE COUTNING OUTPUT ARRAY, then O(n+m)

class Solution {
  public int[][] intervalIntersection(int[][] A, int[][] B) {
    List<int[]> ans = new ArrayList();
    int i = 0, j = 0;

    while (i < A.length && j < B.length) {
      // Let's check if A[i] intersects B[j].
      // lo - the startpoint of the intersection
      // hi - the endpoint of the intersection
      int lo = Math.max(A[i][0], B[j][0]);
      int hi = Math.min(A[i][1], B[j][1]);
      if (lo <= hi)
        ans.add(new int[]{lo, hi});

      // Remove the interval with the smallest endpoint
      if (A[i][1] < B[j][1])
        i++;
      else
        j++;
    }

    return ans.toArray(new int[ans.size()][]);
  }
}